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Question: Show that the circle drawn on a focal chord of a parabola $y^2=4ax$, as a diameter touches the directrix.

Let the parabola be $y^2=4ax$

Let the focal chord be $y = m(x-a) $

Subbing in $y^2=4ax$

$$y^2=4ax$$

$$ \Leftrightarrow (m(x-a))^2 = 4ax $$

$$ \Leftrightarrow m^2 (x^2-2ax+a^2) = 4ax $$

$$\Leftrightarrow m^2x^2 - 2am^2x + m^2a^2 - 4ax = 0 $$

$$ \Leftrightarrow m^2x^2 -(2am^2+4a)x + m^2a^2 = 0 $$

If $x_1$ and $x_2$ are roots then

$$ x_1 + x_2 = \frac{2am^2+4a}{m^2}$$

$$ \therefore x_1 + x_2 = 2a + \frac{4a}{m^2} $$

and $$ x_1 \cdot x_2 = \frac{m^2a^2}{m^2} $$

$$ \therefore x_1 \cdot x_2 = a^2 $$

Corresponding

$$y_1 + y_2 = m(x_1 - a + x_2 - a)$$

$$y_1 + y_2 = m(x_1 + x_2 - 2a)$$

$$y_1 + y_2 = m(2a + \frac{4a}{m^2} - 2a)$$

$$ \therefore y_1 + y_2 = \frac{4a}{m} $$

$$ y_1 \cdot y_2 = m^2(x_1-a)(x_2-a) $$

$$y_1 \cdot y_2 = m^2(x_1x_2 - a(x_1+x_2) + a^2) $$

$$ y_1 \cdot y_2 = m^2( a^2 - a^2(2 + \frac{4}{m^2}) + a^2) $$

$$ y_1 \cdot y_2 = m^2 (\frac{-4a^2}{m^2}) $$

$$ y_1 \cdot y_2 = -4a^2 $$

Now consider

$$ (x_1 - x_2)^2 = (x_1+x_2)^2 - 4x_1x_2 $$

$$ (x_1 - x_2)^2 = a^2(2 + \frac{4}{m^2})^2 - 4a^2 $$

$$ (x_1 - x_2)^2 = a^2(4+\frac{16}{m^2} + \frac{16}{m^4}) - 4a^2 $$

$$ (x_1 - x_2)^2= \frac{16a^2}{m^2} + \frac{16a^2}{m^4} $$

and

$$ (y_1 - y_2)^2 = (y_1+y_2)^2 - 4y_1y_2 $$

$$(y_1 - y_2)^2 = (\frac{4a}{m})^2 -4 \cdot -4a^2 $$

$$(y_1 - y_2)^2 = \frac{16a^2}{m^2} + 16a^2 $$

Therefore

$$ (x_1 - x_2)^2 + (y_1 - y_2)^2 = \frac{16a^2}{m^2} + \frac{16a^2}{m^4} + \frac{16a^2}{m^2} + 16a^2 $$

$$ (x_1 - x_2)^2 + (y_1 - y_2)^2 = 16a^2(\frac{1}{m^4} + \frac{2}{m^2} + 1) $$

$$ (x_1 - x_2)^2 + (y_1 - y_2)^2 = 16a^2(\frac{1}{m^2} + 1)^2 $$

Hence diameter of the circle is given as

$$ D = \sqrt{16a^2(\frac{1}{m^2} + 1)^2} $$

$$ \therefore D = 4a(\frac{1}{m^2} + 1) $$

Now I am stuck how do I show the circle touches the directrix?

I would like to know if my method is correct , and how I could continue with this.

I would also like to see alternative methods (geometrical and other algebraic)! Thank you.

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  • $\begingroup$ I just noticed the very strange situation that your question is a complete duplicate of the same question (math.stackexchange.com/q/1773776) asked a month before by yourself without saying anything... $\endgroup$ – Jean Marie Mar 8 '17 at 8:26
  • $\begingroup$ @JeanMarie Lol this was a year ago , I don't really remember what I was thinking but maybe I was wondering if there were different methods of solving the question (which this post did provide) $\endgroup$ – bigfocalchord Mar 8 '17 at 8:27
  • $\begingroup$ Of course it is an old thing. But you should have made a reference to the first post. One reason is that it avoids people to do again the same thing. Crossed linking is very important. A general comment: such a site will work properly if people act in a respectful manner with others. $\endgroup$ – Jean Marie Mar 8 '17 at 8:33
  • $\begingroup$ @JeanMarie yes I do understand that now as a long user of stackexchange :) back then I was still learning. $\endgroup$ – bigfocalchord Mar 8 '17 at 8:34
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The equation of the circle is given by $$\left(x-\frac{x_1+x_2}{2}\right)^2+\left(y-\frac{y_1+y_2}{2}\right)^2=\frac 14\left((x_1-x_2)^2+(y_1-y_2)^2\right)$$ From what you've got, you can write it as $$\left(x-\left(a+\frac{2a}{m^2}\right)\right)^2+\left(y-\frac{2a}{m}\right)^2=4a^2\left(\frac{1}{m^2}+1\right)^2$$ Now, set $x=-a$ which is the equation of the directrix to have $$\left(y-\frac{2a}{m}\right)^2=0$$ from which the claim follows.

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enter image description here Let the focal chord cut the parabola at $A,A'$ and let $C$ be the midpoint of $A, A'$.

By definition, the distance from both $A, A'$ to the focus $(a,0)$ is equal to the respective distances to the directrix $x=-a$. Let these distances be $m,n$ respectively.

The radius of the circle with $AA'$ as diameter has radius $\frac {m+n}2$, as $AA'=m+n$.

As $C$ is the midpoint of $A,A'$, its distance to the directrix is the average of $m,n$, i.e. $\frac {m+n}2$. Let $D$ = foot of perpendicular from $C$ to the directrix.

Since $CD=CA=CA'$, the circle with the focal chord of the parabola as diameter touches the directrix of the parabola. $\qquad\blacksquare$

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  • $\begingroup$ That is a beautiful graphic , may I ask what program did you make it on? $\endgroup$ – bigfocalchord Jul 2 '16 at 23:42
  • $\begingroup$ Thank you! It's desmos. $\endgroup$ – hypergeometric Jul 3 '16 at 2:16
  • $\begingroup$ Would you happen to have saved the graphing? ,I'm trying to replicate the picture but I am stuck... $\endgroup$ – bigfocalchord Jul 3 '16 at 3:10
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    $\begingroup$ Here it is! $\endgroup$ – hypergeometric Jul 3 '16 at 5:09
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HINT:

Using this OR this, WLOG the endpoints of one diameter the relevant circle can be written as $$(at^2,2at), (a/t^2,-2a/t)$$

Use this, to form the equation of the circle

Now, as the equation of directrix is $x+a=0,$ put $x=-a$ in the equation of the circle

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